The gap between the plates of a parallel plate capacitor is filled with glass of dielectric constant `k=6` and of specific resistivity `100 G Omega m`. The capacitance of the capacitor is `4.0mF`. When a voltage of `2.0 kV` is applied to the capacitor, the leakage current of the capacitor will be.

Options

A. `2.0 xx 10^(-6) A`
B. `3.0 xx 10^(-6) A`
C. `1.5 xx 10^(-3) A`
D. Zero

Answer

C

Explanation

Let `Q_(0)` be the initial charge on the capacitor and `V_(0)` its initial potential. Let `A` be the area of the plates and `l` be the distance between them.
Capacitance `C` of the capacitor is `C = (k epsilon_(0)A)/(l)`
Resistance of the capacitor is `R=(rho l)/(A)=(rho k epsilon_(0))/(C)` ltbegt If `q` be the charge at time `t`, then the leakage current `i` is `i=(dq)/(dt)=(V)/(R)` where `V` is the volt age acros the capacitor at time `t`. But `q = CV`
Hence `i=(dq)/(dt)=C(dV)/(dt)=(V)/(R)`
Hence `(dV)/(V)=-(1)/(CR)dt`. The minus sign is because `dV` is negative.
Integrating between limits, `V=V_(0)exp(-(t)/(CR))`
Hence `i=(V)/(R)=(V_(0))/(R) exp(-(t)/(CR))`
This gives the leakage current `i` at instant `t`. Initial leakage current is obtained at `t =0`
Hence initial leakage current.
`= (i)_(t=0)=(V_(0))/(R)=(Q_(0))/(CR)=(Q_(0))/(rho k epsilon_(0))`
`= (4xx10^(-9)xx2xx10^(3))/(100 xx 10^(9)xx6xx8.85 xx10^(-12))=1.5 xx 10^(-6)A`.